2 (Strong) Exponential Time Hypothesis

Version: 174 In this chapter, we will introduce one of the main hardness assumptions used in this course, the Strong Exponential Time Hypothesis. To this end, we ﬁrst discuss current algorithms for Satisﬁability. We then postulate increasingly strong assumptions about its time complexity: P 6= NP, the Exponential Time Hypothesis (ETH), and the Strong Exponential Time Hypothesis (SETH). To illustrate them, we give conditional lower bounds for the dominating set problem based on these assumptions. Finally, we show that SETH implies the OV hypothesis, making it a stronger assumption.

Let us ﬁrst introduce the satisﬁability problem. Recall that a Boolean formula φ is in conjunctive normal form (CNF) if it is a conjunction of clauses. A clause is a disjunction of literals, and each literal is either a (x1 ∨ x2 ∨ x3) ∧ (x2 ∨ x3 ∨ x4) ∧ (x2 ∨ x4) Boolean variable x or its negation x . i i Figure 2.1: Example of a CNF formula Problem 2.1. Let k ∈ N. k-SAT Given: CNF formula φ with:

• N variables • clause width k • M clauses

Determine: Is there a satisfying assignment for φ? A satisfying assignment for φ is an as- By Cook’s Theorem, we know that k-SAT is NP-hard for k ≥ 3. signment of true/false values to the vari- Thus, P 6= NP implies that there is no polynomial-time algorithm for ables such that all the clauses of φ are satisﬁed, i.e, it contains x where x is set k-SAT for k ≥ 3. (For k = 2, we can solve the problem in polynomial i i to true or a negation xi where xi is set to time.) false. Note: There are at most (N)2k distinct Let us introduce our running example for this chapter: the domi- k clauses of width k. Thus, for ﬁxed k ∈ nating set problem. N, we typically assume without loss of N k k generality that M ≤ ( k )2 = O(N ); i.e., Problem 2.2. the input for k-SAT is of polynomial size in N. 2 fine-grained complexity theory

Dominating Set (DomSet) Given: Undirected graph G = (V, E) with n vertices, integer q ∈ N. Determine: Is there a dominating set S of size q? A dominating set S is a subset n A baseline algorithm solves Dominating Set in time O (q) · qn : of V such that each v ∈ V is For each size-q subset of V, we check whether each v ∈ V satisﬁes dominated by S, i.e., we either v ∈ S or has an edge to some u ∈ S. The latter test can be done in have v ∈ S or there exists an n time O(q) per vertex v, and there are (q) size-q subsets of V, thus the edge {u, v} ∈ E such that u ∈ S. running time bound follows. Unfortunately, as q can be as large as Θ(n), this gives no polynomial-time algorithm in the worst case.

2.1 Lower Bounds under P 6= NP

Could there be a polynomial-time algorithm for Dominating Set? We give a negative answer, assuming P 6= NP. This gives a simple example for a huge number of lower bounds that were proven under the P 6= NP conjecture throughout decades of research since Karps’s list of 21 NP-complete problems.

Theorem 2.3. Dominating Set cannot be solved in polynomial-time unless P = NP.

Proof. We give a standard polynomial-time reduction from the NP- hard problem 3SAT to Dominating Set:

3-SAT Dominating Set

• N variables time O(N + M) • n = 3N + M nodes −−−−−−−−→ • M clauses • q = N

Since the above reduction runs in polynomial time O(N + M), a polynomial-time algorithm for dominating set would allow us to decide any 3SAT instance in polynomial-time, which would contradict P 6= NP. Let us give a reduction as speciﬁed above. Let φ be a 3-CNF formula. We construct a graph G as follows: For each variable xi, we introduce literal vertices xi, xi and a dummy vertex di. We connect these three vertices to form a triangle. Additionally, for each clause Cj, we introduce a clause vertex representing Cj. Finally, we connect a literal .

` C C ` .

vertex to a clause vertex if and only if contains the literal , i.e., ... j j . C = (` ∨ `0 ∨ `00) for some literals `0, `00. Observe that G has 3N + M . .

vertices...... Claim 2.4. G has a dominating set of size N if and only if φ is satisﬁable. .

Figure 2.2: Illustration of constructed graph G. (strong) exponential time hypothesis, version: 174 3

Proof. If there is a satisfying assignment for φ, then let S be the set of literal vertices ` such that ` is true under this assignment. Note that S includes for any variable xi precisely one of xi and xi and thus |S| = N. We show that S is a dominating set. For each variable xi, the vertices xi, xi, di are dominated: one of xi and xi is in S, and the other two vertices are connected to this vertex. Furthermore, each clause vertex Cj is dominated: Our satisfying assignment sets some literal ` of Cj to true. Thus, the literal vertex `, which is contained in S, has an edge to the clause vertex for Cj, which is thus dominated by S. Conversely, let S be a dominating set in G of size N. To dominate all vertices xi, xi, di for i ∈ [N], S must contain, for each i ∈ [N], precisely one of xi, xi, di. We define an assignment for φ, by setting each xi to true if xi ∈ S, to false if xi ∈ S and to an arbitrary value, say true, if di ∈ S. We claim that this assignment satisfies φ: Each clause vertex Cj must be dominated by some literal vertex ` in S, which by definition requires that Cj contains `. As our assignment is chosen such as to set every literal ` ∈ S to true, Cj must be satisfied.

The above claim proves correctness of the reduction. Furthermore, observe that the graph G can be constructed in time O(N + M). Con- sequently, if Dominating Set is solvable in polynomial time, then 3-SAT is solvable in polynomial time.

2.2 Lower Bounds under ETH

Given that we do not expect polynomial-time algorithms for k-SAT and Dominating Set, what are the fastest (superpolynomial) algorithms for these problems? Let us ﬁrst consider this question for 3-SAT. It is hardly surprising Notation: O∗(·) hides polynomial fac- that an extensive line of research has tried to obtain faster and faster tors. algorithms for this problem:

1 Burkhard Monien and Ewald Specken- 3-SAT Algorithms (partial list): meyer. Solving satisﬁability in less than 2n steps. Discrete Applied Mathematics, 10 • O∗ 2N: trivial – check each of the 2N assignments in poly(N) time (3):287–295, 1985 2 O∗ N 1 Uwe Schöning. A probabilistic algo- • 1.6181 : Monien and Speckenmeyer rithm for k-SAT and constraint satisfac- ∗ • O 1.3334N : Schöning2, randomized tion problems. In FOCS’99, pages 410– • O∗ 1.3334N: Moser, Scheder3 414, 1999 3 Robin A. Moser and Dominik Scheder. • ... A full derandomization of Schöning’s k- ∗ • O 1.3280N : Liu4 SAT algorithm. In STOC’11, pages 245– • O∗ 1.3071N: PPSZ algorithm5 with modiﬁcations by Hertli6, ran- 252, 2011 4 Sixue Liu. Chain, generalization of cov- domized ering code, and deterministic algorithm for k-SAT. In ICALP’18, pages 88:1–88:13, A natural question is: When the research effort of the community 2018 5 Ramamohan Paturi, Pavel Pudlák, Michael E. Saks, and Francis Zane. An improved exponential-time algorithm for k-SAT. J. ACM, 52(3):337–364, 2005 6 Timon Hertli. 3-SAT faster and simpler - Unique-SAT bounds for PPSZ hold in general. SIAM J. Comput., 43(2):718–729, 2014 4 fine-grained complexity theory

tends to inﬁnity, what is the fastest algorithm that we will obtain? More formally, we deﬁne the following constant:

Deﬁnition 2.5. For any k ≥ 3, we deﬁne

δN sk := inf{δ | k-SAT has an O(2 )-time algorithm}.

Intuitively, we would hope that if we continue our collective search for faster algorithms, eventually, we will ﬁnd O(2sk·n+ε)-time k-SAT algorithms for some very small ε > 0. Currently, it is far from clear what the true value of, e.g., s3 should be. If P = NP, or√ if 3-SAT has a subexponential algorithm (e.g., run- N ning in time O(2 )), then s3 = 0. The fastest algorithm in the list above gives the upper bound s3 < 0.3863. Given the importance of 3-SAT for theoretical computer science, and the extensive effort of algorithmic research on this problem, it is plausible to conjecture that in fact s3 > 0 – this is in essence the conjecture that 3-SAT has no subexponential-time algorithms. This conjecture is called the Exponential Time Hypothesis postulated by Impagliazzo and Paturi7. 7 Russell Impagliazzo and Ramamohan Paturi. On the complexity of k-SAT. Jour- Hypothesis 2.6 (Exponential Time Hypothesis (ETH)). The Exponential nal of Computer and System Sciences, 62(2): Time Hypothesis postulates that 367–375, 2001

s3 > 0.

An equivalent formulation is the following:

There is some δ > 0 such that 3-SAT cannot be solved in time O(2δN).

A note on randomness: In the deﬁnition of sk (and thus in the hypothesis above), we allow for both deterministic and randomized algorithms. This is sometimes called randomized ETH – however, we will not distinguish between deterministic and randomized ETH. In fact, throughout this course, we will conjecture lower bounds both for deterministic and randomized algorithms (unless speciﬁcally stated otherwise).

Note that this assumption is stronger than P 6= NP. As such, we can actually get stronger lower bounds based on this conjecture (than based on P 6= NP). In fact, the reduction above immediately gives the following lower bound for dominating set:

Theorem 2.7 (Simple ETH lower bound). There is some δ > 0 such that 1/3 Dominating Set cannot be solved in time O(2δn ) unless ETH fails. (strong) exponential time hypothesis, version: 174 5

Proof. Assume that the converse is true, i.e., that for all δ > 0, Domi- 1/3 nating Set has an O(2δn )-time algorithm. Given an arbitrary 3SAT instance φ, we use the reduction of the proof of Theorem 2.3 to construct, in polynomial time, a graph G on n = 3N + M nodes such that φ is satisﬁable if and only if G has a dominating set of size N. Note that N 3 3 we may assume without loss of generality that φ has M ≤ ( 3 )2 ≤ 8N clauses. Thus, n = 3N + M ≤ 11N3. Consequently, constructing G takes time poly(N) and the above Dominating Set algorithm decides 1/3 0 whether G has a dominating set of size N in time O(2δn ) = O(2δ N) √ 0 with δ0 = 3 11 · δ. As this holds for all δ, we obtain a O(2δ N)-time 3SAT algorithm for all δ0 > 0, which would refute the Exponential Time Hypothesis.

1/3 Why do we only obtain a lower bound of 2Ω(n ) instead of 2Ω(n)? The crucial point is that the number of nodes in the Dominating Set instance may get as large as Θ(M) = Θ(N3) instead of Θ(N). However, if we could, for some reason, assume that already 3SAT on a linear number of clauses M = O(N) has no subexponential time algorithms, then we would get a stronger lower bound. Surprisingly, this is possible because of the following result, the Sparsification Lemma due to Impagliazzo, Paturi and Zane8. Let us first state the basic intuition of 8 Russell Impagliazzo, Ramamohan Pa- this result: turi, and Francis Zane. Which problems have strongly exponential complexity? J. Comput. Syst. Sci., 63(4):512–530, 2001 Intuition: Sparsification Lemma The hard case for k-SAT consists of sparse formulas. That is, in reductions from k-SAT, we may assume that M = O(N).

More formally, the lemma describes an algorithm that reduces satisfiability of an N-variable k-CNF formula to a subexponential number of satisfiability instances consisting of sparse N-variable k-CNF formulas (i.e., formulas with M = O(N) clauses). Theorem 2.8 (Sparsification Lemma). Let k ∈ N and ε > 0. There is a constant C = C(k, ε) and an algorithm such that

1. given a k-CNF formula φ, the algorithm computes formulas φ1,..., φt,

2. φ is satisﬁable if and only if there is some i ∈ [t] such that φi is satisﬁable, 3. we have t ≤ 2εN and the algorithm runs in time O(2εNpoly(N)), and

4. each φi is a k-CNF formula with N variables and Mi ≤ C · N clauses. The proof of this result is beyond the scope of this course. Let us use the Sparsiﬁcation Lemma to obtain a stronger lower bound for Dominating Set under ETH. Theorem 2.9. There is some δ > 0 such that Dominating Set cannot be solved in time O(2δn). 6 fine-grained complexity theory

Proof. Let δ > 0, ε := δ/2 and set

δ δ0 := . 2(3 + C(3, ε))

0 Assume there is an algorithm A solving dominating set in time O(2δ n). εN Given a 3-CNF formula φ, we compute φ1,..., φt with t ≤ 2 as in the Sparsiﬁcation Lemma. For each φi, we construct Gi using the reduction in the proof of Theorem 2.3. Note that Gi has ni = 3N + Mi many nodes where Mi ≤ C(3, ε) · N. Thus, running A on Gi takes time

0 0 O 2δ (3N+Mi) = O 2δ (3+C(3,ε))N = O 2δN/2 .

Thus, the total running time to decide satisﬁability of all φi, and thus the satisﬁability of φ, is bounded by t ·O 2δN/2 = O 2(ε+δ/2)N = O 2δN .

0 Thus, if there is an O(2δ N)-time dominating set algorithm for all δ0 > 0, we obtain an O(2δN)-time algorithm for 3-SAT for all δ > 0, contradicting ETH.

In summary, with the above conditional lower bound, we obtain the following state of the art:

1. An O(1.4969n)-time algorithm solves dominating set.9 9 Johan M. M. van Rooij and Hans L. 2. Dominating Set requires time Ω(cn) for some constant c > 0, assum- Bodlaender. Exact algorithms for dominating set. Discrete Applied Mathematics, ing ETH. 159(17):2147–2164, 2011

Let us turn to the question what ETH might say about the polynomial- time regime. In particular consider the problem of ﬁnding small (i.e., constant-sized) dominating sets.

Problem 2.10. Let q ∈ N.

q-Dominating Set Given: Undirected graph G = (V, E) Determine: Does G have a dominating set of size q?

Note that for any ﬁxed q, the problem q-Dominating Set is solv- n able in polynomial time: The O (q) · qn -time algorithm runs in time O(nq+1) for any constant q. Interestingly, for every q ≥ 7, there is + ( ) even an nq o 1 -time algorithm for q-Dominating Set10,11, shaving off 10 Friedrich Eisenbrand and Fabrizio an almost linear factor. Grandoni. On the complexity of ﬁxed parameter clique and dominating set. Theor. Comput. Sci., 326(1-3):57–67, 2004 q+o(1) Let us investigate whether a running time of n should or should 11 √ Mihai Patrascu and Ryan Williams. not be essentially best possible. Could there be a O(n q)-time algo- On the possibility of faster SAT algorithm? Maybe such algorithms are not possible, but a O(nq−2)-time rithms. In Proc. of the 21st Annual ACM- √ SIAM Symposium on Discrete Algorithms q algorithm exists? We will see that ETH implies that no O(n )-time (SODA’10), pages 1065–1075, 2010 (strong) exponential time hypothesis, version: 174 7

algorithms exists for some sufﬁciently large q. For the second question, however, we will not obtain an answer based on the ETH; instead this will be a task for a stronger variant of ETH.

Theorem 2.11. Assuming the ETH, there is some δ > 0 such that q- Dominating Set has no O(nδq)-time algorithms for all sufﬁciently large q.

Proof. We will generalize our reduction from Theorem 2.3 to the following reduction:

k-SAT q-Dominating Set • N variables time O(22N/q) −−−−−−−−→ • n = q2N/q + q + M nodes • M clauses

Note: This reduction has an exponen- > tial blow-up: the number of nodes con- Given this reduction, it sufﬁces to show that for any δ 0, an structed is Θ(2N/q). This is indeed nec- δq O(n )-time algorithm for q-Dominating set for some q ≥ 2/δ would essary to connect the exponential time give a O(2δN)-time algorithm for k-SAT, and thus in particular, 3-SAT12. regime of ETH to the polynomial-time regime of our desired lower bound. = N/q + + To see this, note that the resulting instance consists of n q2 q 12 Note that we have given a general re- M = O(2N/q) nodes – here, we use that M = O(NK) = O(2N/q), as duction from k-SAT, although we only q is a ﬁxed constant. Thus, running the O(nδq)-time algorithm on this need a reduction from 3-SAT at this point – we will make use of the more instance takes time general reduction in the next section. δq O 2N/q = O 2δN .

The reduction time is bounded by O(22N/q) = O(2δN) using q ≥ 2/δ. Thus, we obtain an O(2δN)-time algorithm for 3-SAT, and the claim follows. It remains to design the above reduction. Consider the following basic idea: In the reduction for Theorem 2.3, each vertex of the dominating set “chose” an assignment of true or false for some variable. In our setting, the dominating set has a much smaller size of q – thus, each vertex in the dominating set needs to choose an assignment for a larger number of variables, specifically, for N/q variables. Formally, let φ be a given k-CNF formula and construct a graph G as follows. We partition the variables into q groups G(1),..., G(q), (i) where G = {x(i−1)N/q+1,..., xiN/q}. For each i ∈ [q] and every possible assignment of true/false values to the variables in group G(i), we (i) N/q define a corresponding partial assignment vertex ap , where p ∈ [2 ] (as there are precisely N/q variables in each group). Additionally, for each group G(i), we define a dummy vertex d(i). For each i ∈ [q], (i) (i) we connect all nodes a ,..., a , d(i) to each other; i.e., the vertices 1 2N/q corresponding to a group G(i) form a clique. Furthermore, we introduce a clause vertex for each clause Cj. We (i) connect a partial assignment vertex ap to a clause vertex Cj if and 8 fine-grained complexity theory

(i) only if Cj contains a literal ` ∈ {xk, xk} for some variable xk ∈ G and (i) the assignment represented by ap sets this literal to true.

Figure 2.3: Example for the construction The following claim is analogous to the corresponding claim in the for a 6-variable formula (x1 ∨ x2 ∨ x4) ∧ (x4 ∨ x5 ∨ x6) and q = 3. We label each proof of Theorem 2.3: partial assignment vertex by the set of literals that are true under this assign- Claim 2.12. φ is satisﬁable if and only if G has a dominating set of size q. ment. This claim establishes correctness of the reduction. Finally, observe that we can construct this graph in linear time in the number of its edges. Since this number is bounded by n2 = O(22N/q), the claimed reduction follows.

The above lower bound shows, e.g., that there cannot be an algo- √ rithm that solves q-Dominating Set for arbitrary q in time O(n q), unless the ETH fails. However, if we fix some q, the above result gives no specific lower bound, since any lower bound only applies for suffi- ciently large q. In this sense, this statement might best be described as a statement about the family of problems q-Dominating Set for q ∈ N. In the next section, we will introduce a stronger variant of ETH that allows us to derive conditional lower bounds for specific polynomial- time problems.

2.3 Lower Bounds under SETH

So far, we have discussed speciﬁcally the fastest algorithms for 3-SAT. What are the fastest algorithms for k-SAT with k > 3? Currently, we have the following upper bounds13,14: 13 Ramamohan Paturi, Pavel Pudlák, Michael E. Saks, and Francis Zane. • 3-SAT: O(1.3071N), An improved exponential-time algorithm for k-SAT. J. ACM, 52(3):337–364, 2005 14 Timon Hertli. 3-SAT faster and simpler - Unique-SAT bounds for PPSZ hold in general. SIAM J. Comput., 43(2):718–729, 2014 (strong) exponential time hypothesis, version: 174 9

• 4-SAT: O(1.4690N), •... • k-SAT: O(2(1−c/k)N) for some constant c > 0.

Note that for the current state of the art, the larger k, the more time it takes to solve the problem. In particular, if we let k tend to inﬁnity, the running time of the fastest known k-SAT algorithm approaches O(2N). The Strong Exponential Time Hypothesis postulated by Impagliazzo and Paturi15 states that this is true for the best possible k-SAT algo- 15 Russell Impagliazzo and Ramamohan rithms. Intuitively, one may think of it as follows. Paturi. On the complexity of k-SAT. Jour- nal of Computer and System Sciences, 62(2): 367–375, 2001 Intuition: Strong Exponetial Time Hypothesis For sufﬁcently large k, k-SAT has no O(1.999N)-time algorithms.

Here, we could replace 1.999 by an arbitrary value c < 2. Formally, the hypothesis is stated as follows.

Hypothesis 2.13 (Strong Exponential Time Hypothesis (SETH)). The Strong Exponential Time Hypothesis postulates that

lim sk = 1. k→∞ An equivalent formulation is the following:

For all ε > 0, there is some k ≥ 3 such that k-SAT cannot be solved in time O(2(1−ε)N).

We often state SETH slightly weaker as follows: For no ε > 0, there is an algorithm that solves any k-SAT problem in time O(2(1−ε)N). What evidence is there for SETH? Impagliazzo and Paturi proved that if ETH is true, then sk increases inﬁnitely often. The hypothesized value limk→∞ sk = 1 would be consistent with the current state of the art of k-SAT algorithms. Cygan et al.16 proved SETH is in fact 16 Marek Cygan, Holger Dell, Daniel equivalent to analogous conjectures for other problems such as Hitting Lokshtanov, Dániel Marx, Jesper Ned- erlof, Yoshio Okamoto, Ramamohan Set, Set Splitting and k-NotAllEqual SAT. Paturi, Saket Saurabh, and Magnus Wahlström. On problems as hard as What consequences does SETH imply for polynomial-time algo- CNF-SAT. ACM Trans. Algorithms, 12(3): 41:1–41:24, 2016 rithms? Using the reduction of Theorem 2.11, we obtain the following Note: This list of evidence for SETH is result. not exhaustive! Theorem 2.14. Let q ≥ 3 and ε > 0. There is no q-Dominating Set algorithm running in time O(nq−ε) unless SETH fails.

Proof. Let 0 < ε ≤ 1 and assume that there exists an O(nq−ε)-time q-Dominating Set algorithm. Given a k-SAT instance φ, we use the reduction of the proof of The- orem 2.11 to produce a q-Dominating set instance on n = O(2N/q) 10 fine-grained complexity theory

nodes in time O(22N/q). Running the O(nq−ε)-time q-Dominating Set algorithm on this instance takes time O((2N/q)q−ε) = O(2(1−ε/q)N). Thus, in total, we can decide satisﬁability of φ in time

0 O(22N/q) + O(2(1−ε/q)N) = O(2(1−ε )N), where we set ε0 := ε/q ≤ 1/3. Since k can be chosen arbitrarily, this would show that k-SAT has a (1−ε0)N 0 O(2 )-time algorithm for all k, proving limk→∞ sk ≤ 1 − ε , which would contradict SETH.

The above lower bound yields a very strong conditional lower bound for polynomial-time problems. In particular, already a O(n2.9999)-time algorithm for 3-Dominating Set would refute SETH! Finally, we turn to the connection between SETH and OVH. The following conjecture establishes that OVH is even more plausible than SETH. This greatly helped to popularize OVH as a fundamental hardness assumption for polynomial-time problem. The following reduction is due to Williams17. 17 Ryan Williams. A new algorithm for optimal 2-constraint satisfaction and its Theorem 2.15 (SETH implies OVH). If SETH holds, then for no ε > 0 implications. Theor. Comput. Sci., 348(2- there is an O(n2−εpoly(d))-time OV algorithm. 3):357–365, 2005

Proof. We show the following reduction.

k-SAT OV

N/2 • N variables time O(2N/2 M) • n = 2 vectors −−−−−−−−→ • M clauses • d = M dimensions

Assume that there is a O(n2−εpoly(d))-time OV algorithm for some ε < 1/2. Using this algorithm on the output of the above reduction, we obtain a k-SAT algorithm running in time

0 O(2N/2 M + 2N/2(2−ε)poly(M)) = O(2N(1−ε/2)poly(M)) = O(2(1−ε )N),

for an arbitrary 0 < ε0 < ε/2. Here, we used that M ≤ Nk is polyno- mially small. Thus, as k was chosen arbitrarily, we obtain a contradiction to SETH, as this would give some ε0 > 0 such that k-SAT is solvable in time 0 O(2(1−ε ) N) for all k ≥ 3. It remains to design the above reduction, which is very similar to the reduction from Theorem 2.11: We partition the variables into two

halves, i.e., x1,..., xN/2 and xN/2+1,...,N (here we assume for conve- nience that N is even). Let U denote the set of assignments of Boolean

values to x1,..., xN/2, and similarly, let V contain all assignments to the variables xN/2+1,..., xN. These assignments are called partial assignments. Note that |U| = |V| = 2N/2. (strong) exponential time hypothesis, version: 174 11

For a partial assignment u ∈ U and a clause C, we define sat(u, C) := 1 if and only if u satisfies the clause C, i.e., C contains a literal xi such that xi is set to true in u or C contains a literal xi such that xi is set to false in u. Otherwise, we set sat(u, C) := 0. We define sat(v, C) analogously for v ∈ V. Given partial assignments u ∈ U, v ∈ V, we define the M-dimensional vector

vec(u) := (1 − sat(u, C1), . . . , 1 − sat(u, CM))

vec(v) := (1 − sat(v, C1), . . . , 1 − sat(v, CM))

It is straightforward to verify following claim.

Claim 2.16. For any u ∈ U, v ∈ V, we have that vec(u) and vec(v) are orthogonal if and only if (u, v) gives a satisfying assignment for φ.

Proof. Note that vec(u), vec(b) have an inner product of 0 if and only if for all k = 1, . . . , M, we have sat(u, Ck) = 1, sat(v, Ck) = 1, or both. By deﬁnition, this occurs if and only if each clause Ck is satisﬁed by at least one literal given by the assignment (u, v).

Thus, we can conclude the reduction by deﬁning the vector sets A := {vec(u) | u ∈ U} and B := {vec(v) | v ∈ V}, which are sets of n = 2N/2 vectors in {0, 1}d with d = M, as desired.

The above reduction easily generalizes to an analogous reduction for the so called k-OV problem:

Problem 2.17. k-Orthogonal Vectors (k-OV) d Given: vectors sets A1,..., Ak ⊆ {0, 1} , |A1| = |A2| = ··· = |Ak| = n Determine: Is there a tuple (a1,..., ak) such that for all j ∈ [d] we have a1[j] · a2[j] ··· ak[j] = 0?

Analogously to OV, we can solve k-OV in time O(nkd), but for no k, we know of a O(nk−εpoly(d))-time algorithm. Indeed, using a similar reduction as above, we can show that such algorithms do not exist under SETH. (→ exc.)

In conclusion, we have discussed the conjectures P 6= NP, ETH and P 6= NP ⇑ o(q) SETH. We have seen that while ETH rules out n -algorithms for ETH q-Dominating Set, we need the stronger assumption SETH to prove ⇑ SETH =⇒ OVH tight lower bounds for speciﬁc polynomial-time problems such as 3- Dominating Set or OV (for these, we have proven matching n3−o(1) and n2−o(1)-time lower bounds, respectively). 12 fine-grained complexity theory

Exercises

2.1) Show that Subset Sum has no 2o(n) time algorithm under ETH. 2.2) Show that SETH implies ETH. 2.3) Show that SETH implies that for no k ≥ 2 and ε > 0, there is a O(nk−εpoly(d))-time algorithm for k-OV. Bibliography

Version: 174 Marek Cygan, Holger Dell, Daniel Lokshtanov, Dániel Marx, Jesper Nederlof, Yoshio Okamoto, Ramamohan Paturi, Saket Saurabh, and Magnus Wahlström. On problems as hard as CNF-SAT. ACM Trans. Algorithms, 12(3):41:1–41:24, 2016.

Friedrich Eisenbrand and Fabrizio Grandoni. On the complexity of ﬁxed parameter clique and dominating set. Theor. Comput. Sci., 326 (1-3):57–67, 2004.

Timon Hertli. 3-SAT faster and simpler - Unique-SAT bounds for PPSZ hold in general. SIAM J. Comput., 43(2):718–729, 2014.

Russell Impagliazzo and Ramamohan Paturi. On the complexity of k-SAT. Journal of Computer and System Sciences, 62(2):367–375, 2001.

Russell Impagliazzo, Ramamohan Paturi, and Francis Zane. Which problems have strongly exponential complexity? J. Comput. Syst. Sci., 63(4):512–530, 2001.

Sixue Liu. Chain, generalization of covering code, and deterministic algorithm for k-SAT. In ICALP’18, pages 88:1–88:13, 2018.

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